Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;
int n,x=1;
double dp[2][6001];
int main()
{
	cin>>n;
	dp[0][n]=1;
	for(int i=0;i<n;i++)
	{
		double p;cin>>p;
		for(int j=0;j<=2*n;j++)dp[x][j]=0;
		for(int j=0;j<2*n;j++)dp[x][j+1]+=dp[!x][j]*p;
		for(int j=0;j++<2*n;)dp[x][j-1]+=dp[!x][j]*(1-p);
		x=!x;
	}
	double ans=0;
	for(int i=0;i++<n;)ans+=dp[!x][i+n];
	cout<<fixed<<setprecision(15)<<ans<<endl;
}